New answers tagged

0

Write $f=g\circ \pi + h\circ(1-\pi)$. Wlog assume $\|f\|=\|g\|=1$. Let $x \in M^\perp$, then $f(x)=h(x)=:h$. Suppose this is $>0$ and let $\|x\|=1$. Let $\epsilon>0$, there exists a $y \in M$, $\|y\|=1$ so that $f(y)≥1-\epsilon$. Note $\|\alpha x+ \beta y\|^2=|\alpha|^2 \|x\|^2+|\beta|^2\|y\|^2$, so if $(\alpha,\beta) \in S^2$ you have that $\|\alpha ...


0

To show that the cube is convex, what happens with $\lambda x + (1- \lambda)y$ if $x,y \in [0,a]$ and $\lambda \in [0,1]$? For compactness, as $[0,\frac{1}{n}]$ is compact for all $n \in \mathbb{N}$, the space $X=[0,1] \times [0, \frac{1}{2}] \times [0,\frac{1}{3} ] \times \cdots$ is compact in the product topology. Then you can construct a continuous ...


1

Suppose $T+\mathbb 1$ is invertible and $\|(T-\mathbb 1)(T+\mathbb 1)^{-1}\|≤1$. Then you have: $$(T^\mathstrut-\mathbb 1)(T^*-\mathbb 1)=T^\mathstrut T^* - T^* - T^\mathstrut +\mathbb 1 =(T\mathstrut+\mathbb 1)(T^*+\mathbb 1)-2(T^\mathstrut+T^*)$$ And then you get $$2(T\mathstrut+T^*)=(T^\mathstrut+\mathbb 1)(T^*+\mathbb 1)-(T^\mathstrut-\mathbb ...


3

Then the cardinality is also $\mathfrak{c}$. To see this, note that every element may be written as $\sum_{i\in I}a_i e_i$ where $a_i$ are complex numbers such that at most countably many of them are non-zero and $e_i$ are elements of a fixed orthonormal basis. Then $|H|\leqslant \mathfrak{c}\cdot \mathfrak{c} = \mathfrak{c}$. On the other hand, it is clear ...


0

One way is by contradiction: suppose that for all $n\in\mathbb N$, there exists $u_n\in H^1(\Omega)$ such that $$\|u_n\|_{L^2(\Omega)}> n\|\nabla u_n\|_{L^2(\Omega)}+n\|u_n\|_{L^2(\partial\Omega)}.$$ Then $\|u_n\|_{L^2(\Omega)}>0$ for all $n$, and if we set $v_n=u_n/\|u_n\|_{L^2(\Omega)}$, then $\|v_n\|_{L^2(\Omega)}=1$ for all $n$, and also ...


0

For the further questions: 1. You are right about the meaning of $dt/t$. The factor $(2\pi)^{-1/2}$ appears in order to make $(2\pi)^{-1/2}M$ unitary. In order to show that for each $f\in S$, $$\lVert f\rVert_{L^2(\mathbb R^+,dt/t)}=(2\pi)^{-1/2} \lVert Mf\rVert_{L^2(\mathbb R)},$$ we can do the substitution $t=e^u$ in the integral defining the Mellin ...


1

As Mathematician42 pointed out, the spaces you constructed are not nested. A typical approach to this problem is to use the fact that $H$ is isomorphic to $L^2([0,1])$. Then $E_s=\{f\in L^2 : f(x)=0 \ \forall x>s\}$ works. If you don't want to move to the space of functions, put the basis elements in bijection with the vertices of an infinite binary ...


0

The Closed Graph Theorem shows easily that $T$ is bounded. Hint for the second question: There's a result you know very well that says that if the sequence $(a_n)$ is a ... then $(a_n)$ is weakly square summable. (The statement of this result that you've seen does not use that terminology, but that's what it says.)


0

An ideal is linearly closed. A key step in the GNS construction is that if $x,y\in\mathcal N$, then by Cauchy-Schwarz $$\tag{1} \omega((x+y)^*(x+y))=2\text{Re}\,\omega(y^*x)\leq|\omega(y^*x)|\leq\omega(x^*x)^{1/2}\omega(y^*y)^{1/2}=0. $$ So $$ \mathcal N Y+\mathcal N\subset\mathcal N. $$ The inclusion for the $*$ does not hold. But we don't need it: what ...


1

Actually, you have $H_\perp^1 = \{0\}$, the set containing only the zero function. Indeed, suppose that $u \in H^1(\Omega)$ satisfies $\int_\Omega u \, v \, \mathrm{d}x = 0$ for all $v \in H_0^1(\Omega)$. Then, using the density of $H_0^1(\Omega)$ in $L^2(\Omega)$ you get $\int_\Omega u \, v \, \mathrm{d}x = 0$ for all $v \in L^2(\Omega)$ and this shows $u = ...


2

Use the parallelogram law: $$2\|a\|^2 + 2\|b\|^2 = \|a+b\|^2 + \|a-b\|^2.$$ Suppose that $y_1$ and $y_2$ are both elements of $C$ with $\|y_1 - x_0\| \le r$ and $\|y_2 - x_0\| \le r$. Then $$\|(y_1 + y_2) - 2x_0\|^2 + \|y_1 - y_2\|^2 = 2\|y_1 - x_0\|^2 + 2\|y_2 - x_0\|^2 \le 4r^2.$$ Since $C$ is convex you have $\dfrac{y_1 + y_2}{2} \in C$, so that $$4r^2 ...


3

So this is my final answer for any interested party answer and thank you @joy $\|\frac{x+y}{2}\|^2 \leq \|\frac{x+y}{2}\|^2 + \|\frac{x-y}{2}\|^2 = \frac{1}{2}\|x\|^2+\frac{1}{2}\|y\|^2$ and since $x \neq y$: $\|\frac{x+y}{2}\|^2 < \frac{1}{2}\|x\|^2+\frac{1}{2}\|y\|^2=\frac{1}{2}r^2+\frac{1}{2}r^2=r^2$ Hence: $\|\frac{x+y}{2}\| < r$


4

Use paralleogram law for $\frac{x}{2}$ and $\frac{y}{2}$ to obtain $||\frac{x+y}{2}||^2 + ||\frac{x-y}{2}||^2 = \frac{1}{2}||x||^2 + \frac{1}{2}||y||^2$ and so you get $1$ in the right hand side. Since the LHS is a sum of two non-negative terms, you get the desired inequality since $x\neq y$ .


1

Let $f: \mathbb{R}^n \times l_2 (\mathbb{R} ) \to l_2 $ $$ f((x_1 ,x_2 , ..., x_n ) , (y_i )_{n\geq 1} ) = (x_1 ,x_2 , ..., x_n , y_1 , y_2 , ...)$$ then $f$ is injective.


0

For the identification (or related) see Theorem 4.9 in Rudin1991 ("Functional Analysis") However, this identification (and also the closedness of $V$) need not be invoked for the particular result. I assume you are talking about Corollary 2.5 in Girault and Raviart. Namely, the Hahn-Banach theorem provides that any functional $\mathbf{f}\in V'$ that ...


1

Intuition tells us that the continuous image of a one dimensional set should be, somehow, one-dimensional. E.g. if you look at the image of $t\mapsto (t,t)$ in Euclidean space, which is just a line. From a measure theoretic view, it's volume is $=0$ when using a measure adapted to the dimension of the target space (i.e. a measure which relates, e.g., to ...


2

Well, we have that the bilinear functional $\langle K,T\rangle_\mathrm{HS}$ defined via $$ \langle K,T\rangle_\mathrm{HS}:=\int_0^1\int_0^1k(x,y)\overline{t(x,y)}\mathrm dx\mathrm dy, $$ is a scalar product on the space of Hilbert-Schmidt integral operators. It further induces a (the mentioned) norm by acknowledging that $z\cdot \overline z=|z|^2$ $$ \langle ...


0

Q1: The product measure on the Hilbert cube is obtained as the product of normalized Lebesgue measures on the intervals $[0,1/k]$: that is, $\mathcal L^1$ is multiplied by $k$ to make the measure of $[0,1/k]$ equal to $1$. This makes the product a probability measure. Q2: When $X$ is a metric space and $Y\subset X$, the restriction of the metric gives a ...


1

Hint: if $v_1, v_2, \ldots$ is an orthonormal basis for your Hilbert space and $\lambda_1, \lambda_2, \ldots \in \Bbb{C}$ with $|\lambda_1| = |\lambda_2| = \ldots = 1$, what can you say about $\lambda_1v_1, \lambda_2v_2, \ldots$ and about $f(\lambda_iv_i)$?


0

No, this is even fails in dimension $2$. Just try $C = \{(x,y) \in \mathbb R^2 \mid (x-1)^2 + y^2 \le 2 \}$.


-1

Let $R$ be the projection onto the range of $PQ$. Then, by page 116 of Kadison Ringrose' book. $$R:=P-P\wedge (I-Q). $$ Then, $R \sim P\vee (I-Q) - (I-Q)$. Then,  $\tau(R) =\tau(P\vee (I-Q) - (I-Q)) =\tau(Q) - \tau(I- P\vee (I-Q) ) \le \tau(Q)$. Then, $(Q-R)(H)$ is not empty and it is easy to check $Q-R$ is $Q^\perp \wedge P $.


0

Edit: First we recall the definition: $T$ is a trace-class operator if $$||T||_1:=tr((T^*T)^{1/2})<\infty.$$We show that $H\otimes H$ is dense in the trace class, with respect to that norm. Note that the definition is really all we're going to use about the trace class. In particular: It's not clear to the OP why a trace-class operator actually has a ...


1

No, you can't obtain this norm from an inner product. For instance, the parallelogram law doesn't hold for $X=(1,0)$ and $Y=(0,1)$: $$ \|X+Y\|^2+\|X-Y\|^2=(|1|+|1|)^2+(|1|+|-\!1|)^2=8\ne2\|X\|^2+2\|Y\|^2\;. $$


0

Let $E\subseteq H$ be the closure of $Q(P(H))$. Then $Q$ is a bounded linear map from $P(H)$ to $E$ with dense image, and it follows that $\dim E\leq \dim P(H)$ (if $P(H)$ is finite-dimensional this is just linear algebra; if $P(H)$ is infinite-dimensional, note that its dimension is the same as the minimal cardinality of a dense subset, and then the image ...


2

If $(v,w) \ne 0$, then choose $\alpha$ so that $(v-\alpha w)\perp w$, which is to say that $\alpha=\frac{(v,w)}{(w,w)}$. Then $\alpha\ne 0$, which leads to $$ \|v\|^2=\|(v-\alpha w)+\alpha w\|^2=\|v-\alpha w\|^2+\|\alpha w\|^2 > \|v-\alpha w\|^2. $$ Conversely, if $(v,w)=0$, then the following holds for all $\alpha$: $$ \|v-\alpha ...


1

Let $a = t(v,w)$ for some nonzero real number $t$. Then $$2 t|(v,w)|^2 = 2 \Re (\bar a(v,w)) \le |a|^2 \|w\|^2 = t^2 |(v,w)|^2 \|w\|^2.$$ If $(v,w) \not= 0$ then $2t \le t^2 \|w\|^2$ for all nonzero real $t$, which you can quickly show is impossible.


1

The statement is true. More generally, for every $\alpha\in (0,1)$ the metric space $(\mathbb{R}^n, \|x-y\|^\alpha)$ can be isometrically embedded into a Hilbert space (you only need $\alpha=1/2$ here). This is Corollary 4.8 in the book Embeddings and extensions in analysis by Wells and Williams, who attribute this result to John von Neumann. The ...


4

The claim you are trying to prove is the following statement. Proposition. Let $H$ be a separable Hilbert space and let $(T_n)_{n=1}^\infty$ be a sequence of operators in $B(H)$ which converges to some $T$ in the weak operator topology, i.e., $$\langle T_nx, f\rangle \to \langle Tx, f\rangle$$ for all $x,f\in H$. Then there is a sequence ...


3

Everything has been explained in detail in the other answers, except some issues involving matrix elements of non-self adjoint operators that students often get wrong when they've just learned this topic. Suppose we have an operator $A$ acting on a ket $|k>$ yielding the ket $|q>$: $$|q> = A|k>$$. Then the bra vectors $<q|$ and $<k|$ ...


0

First you can show that $M=\left\{a_{2n-1}:n\right\}^\perp$, so $N\cap M=\left\{x\in N:x\perp\alpha_{2n-1}\forall n\right\}$. Now note that $\alpha_{2n-1}$ is orthogonal to all $\beta_k$, $k\neq n$. Now given $x=\sum_k\lambda_k\beta_k$ take the inner product with $\alpha_{2n-1}$ and use the previous facts to conclude that $\lambda_n=0$. Therefore $x=0$.


3

It can be anything greater than $1$, take for instance $$ A=\begin{bmatrix}0&x\\1/x&0\end{bmatrix} $$


4

If $A$ is symmetric (Hermitian), then that's true (assuming you're talking about the induced Euclidean norm). However, in general, $\|A\|$ can be very large. For example, take $$ A=\pmatrix{1&t\\0&-1} $$ with $t$ as large as you want to make it. Note that for any operator on a Hilbert space, we have $\|A^*A\| = \|A\|^2$ (you should have this as a ...


3

Let $H=\ell^2(\mathbb{N})$ and let $\mathcal{A}$ be the set of all self-adjoint elements in the unit ball of $B(H)$. Define $S_n\colon H\to H, S_n \xi(k)=\xi(k+n)$. Then $\|S_n\|=1$, hence $A_n:=\frac 1 2(S_n+S_n^\ast)\in \mathcal{A}$. The adjoint of $S_n$ is given by $$ S_n^\ast\xi(k)=\begin{cases}\xi(k-n)&\colon k\geq n\\0&\colon k<n\end{cases} ...


1

I'm not exactly sure what level of rigour you're aiming for; the problem may at least in part be that you're expecting more rigour than physicists usually care for; but here are two (possibly related) points at which I don't follow your objections: You write that the Euler-Lagrange equation is $f''(x)=U(x)f(x)$ – that's missing the term $Ef(x)$, which ...


7

What helped me understand it was the notion of bra and ket as vectors in Hilbert space. ket $|f\rangle$ denotes a "usual" vector, bra $\langle x|$ a "transposed" one which can be used for projection. Thus $$\langle x | f \rangle = f(x)$$ is simply the projection of $f$ into its (coordinate) representation of the $x$. This becomes more clear when you see ...


3

Often the inner product (also known as a scalar product or dot product) between two vectors is written as $<v,w>$. The bra and ket notation splits this into two parts, $<v|$ and $|w>$. [In coordinates the "bra" $v$ gets a complex conjugate, so that the inner product is positive.] Furthermore, for convenience vectors $v$ are often written as ...


2

I think that exercise is false. Let $A=\frac{d^2}{dx^2}$ on all finite linear combinations of $\{ 1,x,x^2,x^3,\cdots \}$ and let $B=\frac{d^2}{dx^2}$ on all finite linear combinations of $\{ \sin(\pi x),\sin(2\pi x),\sin(3\pi x),\cdots\}$. Both of these operators are densely-defined; and they're both closable because they're both restrictions of the same ...


2

Assuming that $\wedge_\gamma C_\gamma $ is the intersection, ie the inf, and $\vee_\gamma C_\gamma ^\perp$ is the sup, ie the smallest closed subspace containing each $C_\gamma^\perp$, then yes. (The reason I'm not entirely sure is that this has nothing to do with nets, everything is the same for a family $C_\gamma$, directed or not.) Since ...


10

I would not worry too much at first about distinguishing between "bra" and "ket". What is important is that you have an inner product $\langle x | y \rangle$ which is linear in the second coordinate and conjugate linear in the first, as opposed to the Mathematician's inner product $(x,y)$ which is linear the first coordinate and conjugate linear in the ...


28

First, the $bra$c$ket$ notation is simply a convenience invented to greatly simplify, and abstractify the mathematical manipulations being done in quantum mechanics. It is easiest to begin explaining the abstract vector we call the "ket". The ket-vector $|\psi\rangle $ is an abstract vector, it has a certain "size" or "dimension", but without specifying ...


28

I would like to extend Alex' answer, as well as answer your question in the comments: "Is the "bra" basically something like the dot product?" If you have a vector space $V$ over a field $F$, which is, for now, finite dimensional, you can create another vector space, $V^*$, called the dual space of $V$, which consists of linear functionals defined on $V$. ...


37

In short terms, kets are vectors on your Hilbert space, while bras are linear functionals of the kets to the complex plane $$\left|\psi\right>\in \mathcal{H}$$ \begin{split} \left<\phi\right|:\mathcal{H} &\to \mathbb{C}\\ \left|\psi\right> &\mapsto \left<\phi\middle|\psi\right> \end{split} Due to the Riesz-Frechet theorem, a ...


0

In order to show that $f \in W^{1,2}(\Omega)$, we have to show that it possesses weak derivatives. Recall that $v_i$ is $\partial_i f$ (in a weak sense), if $$\int_\Omega f \, \partial_i \varphi \, \mathrm{d}x = - \int_\Omega v_i \, \varphi \, \mathrm{d}x$$ for all $\varphi \in C_0^\infty(\Omega)$. Now, let $\varphi \in C_0^\infty(\Omega)$ be given. We have ...


2

The arguments I know require that $\Omega$ is both cyclic and separating (i.e., also cyclic for the commutant). When you only require that $\Omega$ is also separating, I don't think that $S_0$ even makes sense. For instance, let $\mathcal M=B(\mathcal H)$, $\Omega=e_1$. Consider the (constant) sequence $E_{12}e_1=0$. Then $S_0E_{12}e_1=E_{21}e_1=e_2\ne0$, ...


1

Suppose that $f,f'$ are absolutely continuous on $\mathbb{R}$ with $f,f''\in L^2$. Then $ff''\in L^1$, which means that the following limits exist $$ \int_{0}^{\pm\infty}ff''dt=-f(0)f'(0)+\lim_{x\rightarrow\pm\infty}f(x)f'(x). $$ If either of these limits is non-zero, then the following diverges in magnitude to $\infty$: $$ ...


0

Hints: If $f,f''\in L^2(\Bbb R)$ does it follow that $f'\in L^2$? Look at the Fourier transform. About limits at infinity: If $f$ is absolutely continuous, $f'\in L^2$ and $a<b$ then $$|f(b)-f(a)|=\left|\int_a^bf'(t)\,dt\right|\le(b-a)^{1/2}||f'||_2,$$so $f$ is uniformly continuous.


0

Functions on a finite set would be characterized by a finite set of parameters. The idea of characterizing polynomials of all orders on a non-trivial interval of $\mathbb{R}$ using a finite set of parameters is immediately objectionable. You just know that cannot be right, which rules out the idea for more general functions that include such polynomials on ...


0

This is because $\lvert x_i\rvert=\sqrt{x_i^2}\le\sqrt{\sum_{i=1}^nx_i^2}=\lVert x\rVert_2.$


1

So the whole space of functions from $\mathbb{R}$ to $\mathbb{R}$ really does contain "uncountable vectors". But we usually deal with much smaller subspaces of that (in the sense of both dimension and cardinality). These are still infinite dimensional, though. To see it, notice that if $f_i$ are nonzero functions and have disjoint support then $\sum_{i=1}^n ...


1

$\newcommand{\Reals}{\mathbf{R}}$Here's an indirect approach (not unrelated to littleO's comment): Let $X = \{1, 2, 3, \dots, n\}$ be a finite set of $n$ elements. The set of all real-valued functions on $X$ is $\Reals^{n}$ by viewing an ordered $n$-tuple $(x_{1}, x_{2}, x_{3}, \dots, x_{n})$ as the function $f:X \to \Reals$ defined by $f(i) = x_{i}$ for $i ...



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